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Physics 02, Gravitational and Spring Potential Energy,Work Energy Theorem

So there is one special thing called potential energy that we call conservative force. So write,
relation between conservative force and potential energy. Relation between conservative
force and potential energy. Okay, class, write now. Change in potential energy is equal to
minus work done by conservative force, okay. Now, from here we will create gravitational
force potential energy (1:36) Conservative force work is multiplied by minus 1 then we will
go towards change in potential energy. So this is gravitational potential energy. Let’s say an
object is moving from one point to another point under the gravitational effect. Like see this,
this is earth surface. One object is here with m mass, let’s say it’s height from the ground is
h1. It moves to this location, whose height from the ground is h2 and between these two
points the vertical separation is capital H, okay. Now, we are moving from point 1 to 2, let’s
say along this path. Now, let’s calculate work done by gravity in this particular path. See
carefully, if we apply a little effort then you will see an interesting thing, we will get two
things in this entire derivation. One is constant force work done, special property of work
done by a constant force and second thing we will see develop gravitational potential energy
also, okay. For complete motion we have observed one thing that on this block one constant
gravitational force is applied downwards, right. Force is constant, when we will tell force is
constant when its direction and magnitude are same, okay. Now, see class by using this
idea, I will write gravitational force like this, okay. Now, this is the first point and this is the
second point. From first point to second point we will travel through an arbitrary path or
track, done. Now, what we did, in the whole displacement, I will select in the entire path I
will select displacement ds, okay. Through the track or path which we are moving, we
selected a small displacement, ds. Yes or no? Now, see, our displacement is a vector
quantity. So, see carefully, what I will do is, I will cut this displacement into two pieces. I will
magnify this displacement. I can see a straight line, ds, this is the displacement vector. Now,
I will make the component of this displacement in the direction of force and perpendicular
to the force. In the direction and perpendicular to the direction, we will make the force as
component. ds, perpendicular to force and ds parallel. See, this is a vector quantity, I can
make it component anywhere and anyhow. Now, I know that force is always vertical here.
So, what I did, I made one component vertical and other component horizontal. One
horizontal component and other vertical component, two pieces, okay. I deliberately divided
the displacement into two pieces, one horizontal and other vertical. That means, this
displacement ds can be written as ds perpendicular plus ds parallel. Let it be any arbitrary
path, selected a small displacement on it and divided that small displacement into two
pieces, ds perpendicular and ds parallel, perpendicular to force and parallel to force, okay.
That means, now I will write, work done by my constant force, what will it be – f dot ds? And
now what I will do is, I will write ds as perpendicular and parallel. Tell me one thing, if I make
two pieces of this full integral, ds perpendicular, ds perpendicular means force and angle
between this force, how much it is, 90 degree. That means the work done value is zero. That
means finally the total work done is F dot ds will be parallel. And if it is constant force then
displacement is useful only when it is parallel to force, okay. The displacement which is
perpendicular to the force in that value will be zero. Okay, force value is constant, so I can
keep force outside integration – ds parallel. And ds parallel can be written as delta s parallel,
the total displacement. That means constant force work done can be written directly, if I am
clear that force is constant and the constant force whether it is conservative or not, the
work done can be written directly. How much? F dot product with the displacement which is
parallel to force. That means in direction of the force or opposite direction of the force. And
if I see carefully, I can simplify this plus minus F delta s parallel. Plus minus F delta s parallel,
why did it put plus minus here, so that dot product can be eliminated and if it is parallel then
the angle between them would be zero degree or 180 degree. When will I apply plus, when
they are in the same direction and when will I apply minus, when they are in opposite
direction. Okay, so we have simplified work done by constant force. As long as you know the
force is constant, you can write straight away work done as plus minus F delta s parallel. In
this full.

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